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Aviation History
1932
1932 - 0278.PDF
SUPPLEMENT TO FLIGHT 18 THE AIRCRAFT ENGINEER MARCH 25, 1932 . U = i (P + aTj, + 6T2 + cT, . . )2 X AE Now the loads in the redundancies adjust themselves so that the total work done on the whole structure is a minimum. The tensions T15 T2, T3 may be calculated by partially differentiating and equating to zero. Thus SU ST SU "AS (P + aT1 + 6T2 + cT, .. .) = 0 , = ^AE (P + aTl + 6T2 + cT, . ..) = 0 SU rl ST =^AE(P + «T1 + 6T2 + cT3...) = 0 In applying this method to the incidence wire of a wing structure it is necessary to know the lengths and areas of each member in the structure. Since these values vary considerably according to the size and type of aircraft, it is not possible to give a general solution. An indication of the probable effect of incidence wires in unsymmetrical loading may be found by examining a particular case. For this investigation a biplane of the proportions shown in Fig. 1 will be used. TABLE I. Geometry of Structure. Member L'ght. Lift f Outer bays i — 2-75 Wires 1 Inner bays — ; 2-0 Incidence wires .. 1 • 0 j — T.C.P. incid. wires.. 10 — T.C.P. cross wires .. — 1-0 Drag J Outer bav.. 10 2-75 Wires'! Inner bay.. 10 2-0 20 20 2-0 10 10 — 3-40 2-83 2-24 1-41 1-41 2-93 2-24 — 0-707 0-446 0-707 0-809 0-588 0-341 0-446 0-707 0-938 0-892 0-707 0-892 0-707 0-707 Further assumptions must be made as regards the applied forces on the structure. The simplicity of these assumptions will not affect the validity of the conclu sions drawn. Accordingly, the entire lift " L " will be taken on the front truss and distributed such that the Lift Wires Top Front Spar Top Rear Spar Btm. Front Spar Btm. Rear Spar TABLE II. Loads in Members. Member C.P. forward Front Outer Wire cut Front Inner Wire cut Area x 10-5 f Front outer j Front inner 1 Rear outer (_ Rear inner r Outer bay •< Inner bay L Centre section ( Outer bay X Inner bay (_ Centre section I Outer bay [Inner bay Outer bay Inner bay 0-340 L 0-566 L 0-275 L 0-675 L 0-675 L 0-275 L ] Outer bay y Inner bay- Top Wing Wirw j Centre secti01 Btm. 1 Wing I Outer bay Drag f Inner bay Wires J Incid. J Wires * Outer struts Inner struts [c/s struts 0-142 L 0-170 L 0142 L 0-038 L 0138 L ; 0-275 L I 0-475 L i 0-475 L 0138 L 0-238 L 0-275 L 0147 L 0112 L 0147 L 0-112 L 0 112 L 0 071 L 0-170 L 0-283 1 0138 L 0-138 1 0063 L 0-400 L 0-400 L 0-80 L 0-50 L 015L 0-25 L 1-2 L 1-2 L 1-2 L 10 L 1-0 1 10 L — 1-0 L 0-063 Li 1-OL 0-8L 0-8 L 0-224 L 0-224 L 0-224 L 0 142 L Centre 1 Section ( Front Cross f Rear Wires J 0-13 L 0-20 L 0-20 L 013 L 0-20 L j 0-10 L ! 0-20 L i 0-13 L 0-20 L 0-20 L All struts have been omitted, since the energy stored in them is small compared with that in spars and wires. The areas of members have been fixed as follow :•— Wing Spars (Steel). Top front spar.—Assume the end load stress in the outer bay to be 10 tons per sq. in. Then area "= 0.275 L/22,400 = 1.2 L X 10 5 approx. Top rear spar.—Assujne 80 per cent, of front = 1.0 L x 10 5 approx. 2-o- FIG. I PROPORTIONS OF BIPLANE applied force at joints, A, B, C, D, E and G, J, K, L, Bottom front spar.—Assume 80 per cent, of top front M, is 0.1 L. 1.0 L X 10-5 approx. Before the areas of members can be estimated, the Bottom rear spar.—Assume 80 per cent, of bottom front loads, as found by standard methods, must be calculated. Table II gives the loads for three cases corresponding to centre of pressure forward, normal flight, and two cut wire conditions. The required factor for the last two cases is half that of the normal flight case. 0.8 L X 10-6 approx. For all wires assume that the maximum load in the three cases considered produces a stress of 50 tons per sq. in. Certain of the wires have to be guessed, e.g., the 260 b
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